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人気のある微分積分 >

integral of (sin(a)x)/(sqrt(c+cos(a)x))

解

\int \frac{sin ( a ) x}{c + cos ( a ) x} d x

解

sec^{2} (a) sin (a) (\frac{2}{3} (c + x cos (a))^{\frac{3}{2}} - 2 c c + x cos (a)) + C

解答ステップ

\int \frac{sin ( a ) x}{c + cos ( a ) x} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= sin (a) \cdot \int \frac{x}{c + cos ( a ) x} d x

u置換積分法を適用する

= sin (a) \cdot \int \frac{u - c}{u cos ^{2} ( a )} d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= sin (a) \frac{1}{cos ^{2} ( a )} \cdot \int \frac{u - c}{u} d u

拡張

\frac{u - c}{u} : u - \frac{c}{u}

= sin (a) \frac{1}{cos ^{2} ( a )} \cdot \int u - \frac{c}{u} d u

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= sin (a) \frac{1}{cos ^{2} ( a )} (\int u d u - \int \frac{c}{u} d u)

\int u d u = \frac{2}{3} u^{\frac{3}{2}}

\int \frac{c}{u} d u = 2 c u

= sin (a) \frac{1}{cos ^{2} ( a )} (\frac{2}{3} u^{\frac{3}{2}} - 2 c u)

代用を戻す

u = c + cos (a) x

= sin (a) \frac{1}{cos ^{2} ( a )} (\frac{2}{3} (c + cos (a) x)^{\frac{3}{2}} - 2 c c + cos (a) x)

基本的な三角関数の公式を使用する:

\frac{1}{cos ( x )} = sec (x)

= sec^{2} (a) sin (a) (\frac{2}{3} (c + x cos (a))^{\frac{3}{2}} - 2 c c + x cos (a))

定数を解答に追加する

= sec^{2} (a) sin (a) (\frac{2}{3} (c + x cos (a))^{\frac{3}{2}} - 2 c c + x cos (a)) + C

グラフ

人気の例

integral of 1/7 x^{-6/7}

\int \frac{1}{7} x^{- \frac{6}{7}} d x

tangent y^2+xy+x^2=1,(0,1)

t an g e n t y^{2} + x y + x^{2} = 1, (0, 1)

x(dy)/(dx)=x^{-3}-2y

x \frac{d y}{d x} = x^{- 3} - 2 y

integral from 1 to 4 of 3/x

\int_{1}^{4} \frac{3}{x} d x

inverselaplace 3/(2s^2+3s+1)

in v erse l a pl a ce \frac{3}{2 s ^{2} + 3 s + 1}